Another solution, it's basically the same as v_Enhance's but perhaps a little more motivated. First let's find conditions on $d,a,b$ such that this holds. We can WLOG $d$ is prime. Now, remark that $d \nmid a,b$ unless $d=2$ because otherwise one of $a^n,b^n$ is constant modulo $d$ so they must be $1 \pmod{d}$, thus $d|2$. Thus if $d|a$ or $d|b$ then we must have $d=2$ and one of $a,b$ is odd while the other is even. Now take $d$ as an odd prime where $d \nmid a,b$. We have $a^n + b^n \equiv -1 \pmod{d}$ for all $n$. Take $n = d-1$ to get $d=3$. Thus the only values of $d$ that work are $d=2,3$. Thus it immediately follows $d$ exists for $a,b$ iff one of $d=2$ or $d=3$ works. In other words, $d$ exists iff one of $a,b$ is even while the other is odd, or if $a \equiv b \equiv 1 \pmod{3}$.

We divide the problem into three cases: Case $1$:$a,b$ are both divisible by $d$. Let $a=da_1,b=db_1$;$a_1,b_1$ are not necessarily coprime with $d$. Then $d^n(a_1^n+b_1^n) \equiv -1 \pmod {d} \Rightarrow d|-1$. So $d=-1,1$ which is a contradiction since $d \ge 2$. Case $2$:One of $a,b$ is coprime with $d$. WLOG Let $(b,d)=1,a=da_1$,$a_1$ is not necessary coprime with $d$. Then $-d^na_1^n \equiv b^n+1\pmod {d} \Rightarrow b^n \equiv -1\pmod {d} \forall n \in \mathbb {N}$ So $b \equiv -1\pmod {d},b^2 \equiv -1\pmod {d}$ $\Rightarrow b(b-1) \equiv 0\pmod {d}$ $\Rightarrow -(b-1) \equiv 0\pmod {d}$ $\Rightarrow b \equiv 1\pmod {d}$ Now $b \equiv -1\pmod {d},b \equiv 1\pmod {d} \Rightarrow d|2 \rightarrow d=2$. Case $3$:$a,b$ are both coprime with $d$. $a^{3n}+b^{3n} \equiv -1\pmod {d}$ $\Rightarrow (a^n+b^n)(a^{2n}-a^nb^n+b^{2n}) \equiv -1\pmod {d}$ $\Rightarrow -(-1-a^nb^n) \equiv -1\pmod {d}$ $\Rightarrow a^nb^n \equiv -2\pmod {d} \forall n \in \mathbb {N}$ So $ab \equiv -2\pmod {d},a^2b^2 \equiv -2\pmod {d}$ $\Rightarrow ab(ab-1) \equiv 0\pmod {d} \Rightarrow ab \equiv 1\pmod {d}$ Now $ab \equiv -2 \pmod {d},ab \equiv 1\pmod {d} \Rightarrow d|3 \rightarrow d=3$. Verification: For $d=2$,let $a \equiv 0\pmod {2},b \equiv 1\pmod {2}$[according to the condition in Case $2$] $\Rightarrow 2|a^n+b^n+1 \forall n \in \mathbb {N}$ which indeed satisfies our condition. For $d=3$,we get three more cases according to the condition in Case $3$. $(1):(a,b) \equiv (-1,-1)\pmod {3}$,a contradiction for $n=1$. $(2):(a,b) \equiv (1,-1)\pmod {3}$,a contradiction for $n=1$. $(3):(a,b) \equiv (1,1)\pmod {3}$ Now $a \equiv 1\pmod {3} \rightarrow a^n \equiv 1\pmod {3},b \equiv 1\pmod {3} \rightarrow b^n \equiv 1\pmod {3}$ Hence $3|a^n+b^n+1 \forall n \in \mathbb {N}$ which satisfies our condition as well So these two sets are the possible families of solutions.

WLOG we will use $a$ to use a property about one $a,b$. Let $gcd(a,d)=k$, and let $p$ be a prime $p|k$ such that $p$ is not $2$. Then $b^n+1\equiv 0\pmod{d}$ for all $n$. But we have that $b^n$ cycles residues mod $p$ unless $b\equiv 0,1\pmod{p}$, but both do not work. Thus in this case $gcd(a,d)=1$. We have that $a+b\equiv a^2+b^2\equiv a^3+b^3\pmod{p}$ and $(a+b)^2-a^2-b^2\equiv a^3+b^3-(a+b)(a^2+b^2)\pmod{p}$, giving $a+b\equiv 2\pmod{p}$ for $a,b$ relatively prime to $d$, implying $d=3$. Now, we note that if there is a divisor $p$ in $d$ with $p$ not $2,3$, we use the above argument to achieve contradiction. Thus if $a$ is not relatively prime to $d$, $d=2$. Our solutions are $a$ odd, $b$ even, $d=2$, and $a\equiv b\equiv\pmod{3}$, $d=3$.

My solution- It's easy to check the possiblity of $d=2,3$ when $a$ and $b$ are of opposite parity and when both $a, b$ leave remainder $1$ when divided by $3$ respectively.When $a, b$ are of same parity d must be odd. Now, $d | a+b+1$ ; also $d | a^{2}+b^{2}+1$ ; also $ d | a^{3}+b^{3}+1$ We can write $d | (a+b)^{2}-2ab+1$ That implies $d | 2-2ab$ or $d | ab-1$(since d is odd) We have $d | (a+b)(a^{2}+b^{2}-ab)+1$ That implies $d | 3$ or $d=3$ This leaves us with no other cases.